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Damping an oscillatory motion: friction on an object connected to a spring

Damping oscillatory motion is important in many systems, and the ability to control the damping is even more so. This is generally attained using non-conservative forces such as the friction between surfaces, and viscosity for objects moving through fluids. The following example considers friction. Suppose a 0.200-kg object is connected to a spring as shown in [link] , but there is simple friction between the object and the surface, and the coefficient of friction μ k size 12{μ rSub { size 8{k} } } {} is equal to 0.0800. (a) What is the frictional force between the surfaces? (b) What total distance does the object travel if it is released 0.100 m from equilibrium, starting at v = 0 size 12{v=0} {} ? The force constant of the spring is k = 50 . 0 N/m size 12{k="50" "." 0`"N/m"} {} .

 The given figure (a) shows a spring on a frictionless surface attached to a bar or wall from the left side and on the right side of the spring, there is an object attached with mass m. Its amplitude is given by X, and X is equal to zero at the equilibrium level. Force F is applied to it from the right side, represented by a red arrow pointing toward the left and velocity v is equal to zero. An arrow showing the direction of force is also given alongside this figure as well as with the other four figures. The energy of the object is half k x squared.           In the given figure (b), after force is applied, the object moves to the left, compressing the spring slightly. The displacement of the object from its initial position is indicated by dots. The force F, here is equal to zero and velocity v, is maximum in the negative direction or the left. The energy of the object in this case is half m times negative v-max whole squared.           In the given figure (c), the spring has been compressed the maximum limit, and the amplitude is minus X. Now the force is toward the right, indicated here with a red arrow pointing to the right and the velocity, v, is zero. The energy of the object now is half k times negative x whole squared.           In the given figure (d), the spring is shown released from its compressed position and the object has moved toward the right side to reach the equilibrium level. Here, F is equal to zero, and the velocity, v, is the maximum. The energy of the object becomes half k times v max squared.           In the given figure (e), the spring has been stretched loose to the maximum possible limit and the object has moved to the far right. Now the velocity v, here is equal to zero and the direction of force is toward the left. As shown here, F is equal to zero. The energy of the object in this case is half k times x squared.
The transformation of energy in simple harmonic motion is illustrated for an object attached to a spring on a frictionless surface.

Strategy

This problem requires you to integrate your knowledge of various concepts regarding waves, oscillations, and damping. To solve an integrated concept problem, you must first identify the physical principles involved. Part (a) is about the frictional force. This is a topic involving the application of Newton’s Laws. Part (b) requires an understanding of work and conservation of energy, as well as some understanding of horizontal oscillatory systems.

Now that we have identified the principles we must apply in order to solve the problems, we need to identify the knowns and unknowns for each part of the question, as well as the quantity that is constant in Part (a) and Part (b) of the question.

Solution a

  1. Choose the proper equation: Friction is f = μ k mg size 12{F=μ rSub { size 8{k} } ital "mg"} {} .
  2. Identify the known values.
  3. Enter the known values into the equation:
    f = (0.0800) (0 .200 kg) (9 .80 m / s 2 ) . size 12{f=0 "." "0800" times 0 "." "200"`"kg" times 9 "." 8`"ms" rSup { size 8{"-2"} } } {}
  4. Calculate and convert units: f = 0.157 N . size 12{F=μ rSub { size 8{k} } ital "mg"} {}

Discussion a

The force here is small because the system and the coefficients are small.

Solution b

Identify the known:

  • The system involves elastic potential energy as the spring compresses and expands, friction that is related to the work done, and the kinetic energy as the body speeds up and slows down.
  • Energy is not conserved as the mass oscillates because friction is a non-conservative force.
  • The motion is horizontal, so gravitational potential energy does not need to be considered.
  • Because the motion starts from rest, the energy in the system is initially PE el,i = ( 1 / 2 ) kX 2 size 12{ ital "PE" rSub { size 8{e1} } = \( 1/2 \) ital "kX" rSup { size 8{2} } } {} . This energy is removed by work done by friction W nc = fd size 12{W rSub { size 8{ ital "nc"} } = ital "fd"} {} , where d size 12{x} {} is the total distance traveled and f = μ k mg size 12{f=μk ital "mg"} {} is the force of friction. When the system stops moving, the friction force will balance the force exerted by the spring, so PE e1,f = ( 1 / 2 ) kx 2 size 12{"PE" rSub { size 8{"e1,f"} } = \( 1/2 \) ital "kx" rSup { size 8{2} } } {} where x size 12{x} {} is the final position and is given by
    F el = f kx = μ k mg x = μ k mg k . alignl { stack { size 12{F rSub { size 8{"el"} } =f} {} #ital "kx"=μ rSub { size 8{k} } ital "mg" {} # x= { {μ rSub { size 8{k} } ital "mg"} over {k} } {}} } {}
  1. By equating the work done to the energy removed, solve for the distance d size 12{x} {} .
  2. The work done by the non-conservative forces equals the initial, stored elastic potential energy. Identify the correct equation to use:
    W nc = Δ KE + PE = PE el,f PE el,i = 1 2 k μ k mg k 2 X 2 . size 12{W rSub { size 8{"nc"} } =Δ left ("KE"+"PE" right )="PE" rSub { size 8{"el,f"} } - "PE" rSub { size 8{"el,i"} } = { {1} over {2} } k left ( left ( { {μ rSub { size 8{k} } ital "mg"} over {k} } right ) rSup { size 8{2} } - X rSup { size 8{2} } right )} {}
  3. Recall that W nc = fd size 12{W rSub { size 8{ ital "nc"} } = ital "Fd"} {} .
  4. Enter the friction as f = μ k mg size 12{F=μ rSub { size 8{k} } ital "mg"} {} into W nc = fd size 12{W rSub { size 8{ ital "nc"} } = ital "Fd"} {} , thus
    W nc = μ k mgd . size 12{W size 8{"nc"}=μ rSub { size 8{k} } ital "mgd"} {}
  5. Combine these two equations to find
    1 2 k μ k mg k 2 X 2 = μ k mgd . size 12{ { {1} over {2} } k left ( left ( { {μ rSub { size 8{k} } } over {k} } right ) rSup { size 8{2} } - X rSup { size 8{2} } right )= - μ rSub { size 8{k} } } {}
  6. Solve the equation for d size 12{x} {} :
    d = k 2 μ k mg ( X 2 ( μ k mg k ) 2 ) . size 12{d= { { { {1} over {2} } ital "kX" rSup { size 8{2} } } over {μ rSub { size 8{k} } } } } {}
  7. Enter the known values into the resulting equation:
    d = 50 . 0 N/m 2 0 . 0800 0 . 200 kg 9 . 80 m/s 2 0 . 100 m 2 0 . 0800 0 . 200 kg 9 . 80 m/s 2 50 . 0 N/m 2 . size 12{d= { {"50" "." 0" N/m"} over {2 left (0 "." "0800" right ) left (0 "." "200"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right )} } left ( left (0 "." "100"" m" right ) rSup { size 8{2} } - left ( { { left (0 "." "0800" right ) left (0 "." "200"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right )} over {"50" "." 0" N/m"} } right ) rSup { size 8{2} } right )} {}
  8. Calculate d size 12{x} {} and convert units:
    d = 1 . 59 m . size 12{d=1 "." "59"`m} {}

Discussion b

This is the total distance traveled back and forth across x = 0 size 12{x=0} {} , which is the undamped equilibrium position. The number of oscillations about the equilibrium position will be more than d / X = ( 1 . 59 m ) / ( 0 . 100 m ) = 15 . 9 size 12{d/X= \( 1 "." "59"`m \) / \( 0 "." "100"`m \) ="15" "." 9} {} because the amplitude of the oscillations is decreasing with time. At the end of the motion, this system will not return to x = 0 size 12{x=0} {} for this type of damping force, because static friction will exceed the restoring force. This system is underdamped. In contrast, an overdamped system with a simple constant damping force would not cross the equilibrium position x = 0 size 12{x=0} {} a single time. For example, if this system had a damping force 20 times greater, it would only move 0.0484 m toward the equilibrium position from its original 0.100-m position.

This worked example illustrates how to apply problem-solving strategies to situations that integrate the different concepts you have learned. The first step is to identify the physical principles involved in the problem. The second step is to solve for the unknowns using familiar problem-solving strategies. These are found throughout the text, and many worked examples show how to use them for single topics. In this integrated concepts example, you can see how to apply them across several topics. You will find these techniques useful in applications of physics outside a physics course, such as in your profession, in other science disciplines, and in everyday life.

Why are completely undamped harmonic oscillators so rare?

Friction often comes into play whenever an object is moving. Friction causes damping in a harmonic oscillator.

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Describe the difference between overdamping, underdamping, and critical damping.

An overdamped system moves slowly toward equilibrium. An underdamped system moves quickly to equilibrium, but will oscillate about the equilibrium point as it does so. A critically damped system moves as quickly as possible toward equilibrium without oscillating about the equilibrium.

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Section summary

  • Damped harmonic oscillators have non-conservative forces that dissipate their energy.
  • Critical damping returns the system to equilibrium as fast as possible without overshooting.
  • An underdamped system will oscillate through the equilibrium position.
  • An overdamped system moves more slowly toward equilibrium than one that is critically damped.

Conceptual questions

Give an example of a damped harmonic oscillator. (They are more common than undamped or simple harmonic oscillators.)

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How would a car bounce after a bump under each of these conditions?

  • overdamping
  • underdamping
  • critical damping

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Most harmonic oscillators are damped and, if undriven, eventually come to a stop. How is this observation related to the second law of thermodynamics?

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Problems&Exercises

The amplitude of a lightly damped oscillator decreases by 3 . 0% size 12{3 "." 0%} {} during each cycle. What percentage of the mechanical energy of the oscillator is lost in each cycle?

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Questions & Answers

it is a process in which electrolyte breaks down into its ions....
ABHISHEK Reply
ionization
Shang
A ball is thrown vertically upwards so that its max height is 100m...at the same instant another ball is dropped frm 100m above the ground...where will the two balls meet and why?
Blessing Reply
50m
wenhe
why?
Blessing
plz calculate the answer...
Blessing
😟
Blessing
75m
S.M
how?😨
Blessing
because they approach each other at different speeds, but same acceleration, so you can calculate its height above ground.
S.M
Yeah. He's right. Your question seems very similar to a question I did in physics class, and our answer was 50.so yeah that's why I said that. But 75m is correct
wenhe
what's is waves
nuraddeen
Waves is the disturbance of a medium such that the particles in the medium vibrate to cause the movement of energy from one point to another without the the particle itself moving.
Prince
what is the medium that light waves travel through?
Luke
and electrons, protons, neutrons and quarks for that matter?
Luke
it's spacetime that light and every other particles travel through actually
S.M
ether
Anand
for above question, i think 50m is the correct answer. the first ball will behave exactly like 2nd ball when it reaches 100m (max height).
James
From rest, a body moves with an acceleration of 8m/s in 10 secs. calculate the distance during the 8th seconds
Caleb
please use energy conservation to solve this problem. i think i have an answer but it is too long to explain here.
James
one is with some initial velocity v, while other one will start it's journey from rest, so they can't travel equal distance in equal amount of time
S.M
to be honest I knew the answer was just pointing out a flaw in above answer and light does not require a medium to travel through, this was proved in the 1887 by Michelson and Morley.
Luke
the two balls have different acceleration. the accelerations are same in magnitude but in different directions for them. so the first ball will gradually move slower and the 2nd ball will move faster.
James
u can just put 50m or 75m in equation to check whether right or not
James
the total energy of the two balls are equal, KE + PE. 1/2mv(1)^ 2 + mgh(1) = 1/2mv(2)^2 +mgh(2). use that
James
what's is electrolysis
nuraddeen
Give an example (but not one from the text) of a device used to measure time and identify what change in that device indicates a change in time.
David Reply
hour glass, pendulum clock, atomic clock?
S.M
tnks
David
A heart pumping blood would indicate a change in time as its volume or pressure changed. The ratio of displacement or change in configuration between any 2 systems can indicate time.
Khashon
how did they solve for "t" after getting 67.6=.5(Voy + 0)t
Martin Reply
Find the following for path D in [link] : (a) The distance traveled. (b) The magnitude of the displacement from start to finish. (c) The displacement from start to finish.
David Reply
the topic is kinematics
David
can i get notes of solid state physics
Lohitha
just check the chpt. 13 kinetic theory of matter it's there
David
is acceleration a fundamental unit.
David Reply
no it is derived
Abdul
no
Nisha
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David
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Emmanuel
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Gift
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Emmanuel
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Villaflor Reply
how about a conceptual framework can you simplify for me? needed please
Villaflor
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Afzal
I think they are constantly moving
Villaflor
yep what is problem you are stuck into context?
S.M
not possible to fix electron position in space,
S.M
Physics
Beatriz
yes of course Villa flor
David
equations of kinematics for constant acceleration
Sagcurse Reply
A bottle full of water weighs 45g when full of mercury,it weighs 360g.if the empty bottle weighs 20g.calculate the relative density of mercury and the density of mercury....pls I need help
Lila Reply
well You know the density of water is 1000kg/m^3.And formula for density is density=mass/volume Then we must calculate volume of bottle and mass of mercury: Volume of bottle is (45-20)/1000000=1/40000 mass of mercury is:(360-20)/1000 kg density of mercury:(340/1000):1/50000=(340•40000):1000=13600
Sobirjon
the latter is true
Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture...take density of water as 1g/cm3 and density of liquid 1.2g/cm3
Lila
plz hu can explain Heisenberg's uncertainty principle
Emmanuel Reply
who can help me with my problem about acceleration?
Vann Reply
ok
Nicholas
how to solve this... a car is heading north then smoothly made a westward turn during the travel the speed of the car remains constant at 1.5km/h what is the acceleration of the car? the total travel time of the car as it smoothly changed its direction is 15 minutes
Vann
i think the acceleration is 0 since the car does not change its speed unless there are other conditions
Ben
yes I have to agree, the key phrase is, "the speed of the car remains constant...," all other information is not needed to conclude that acceleration remains at 0 during the entire time
Luis
who can help me with a relative density question
Lila
1cm3 sample of tin lead alloy has mass 8.5g.the relative density of tin is 7.3 and that of lead is 11.3.calculate the percentage by weight of tin in the alloy. assuming that there is no change of volume when the metals formed the alloy
Lila
So it looks like you need a formula for rotational acceleration. Are you asking about its angular acceleration?
Khashon
morning, what will happen to the volume of an ice block when heat is added from -200°c to 0°c... Will it volume increase or decrease?
adefenwa Reply
no
Emmanuel
hi what is physical education?
Kate
BPED..is my course.
Kate
No
Emmanuel
I think it is neither decreases nor increases ,it remains in the same volume because of its crystal structure
Sobirjon
100g of water is mixed with 60g of a liquid of relative density 1.2.assuming no changes in volume occurred,find the average relative density of the mixture. take density of water as 1g/cm3 and density of liquid as 1.2g/cm3
Lila
Sorry what does it means"no changes in volume occured"?
Sobirjon
volume can be the amount of space occupied by an object. But when an object does not change in shape it will still occupy the same space. Thats why the volume will still remain the same
Ben
Most soilds expand when heated but if it changes state at 0C it will have less volume. Ice floats because it is less dense ie a larger mass per unit volume.
Richard
how to calculate velocity
Okwethu Reply
v=d/t
Emeka
his about the speed?
Villaflor
how about speed
Villaflor
v=d/t
Nisha
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lawan Reply
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akinmeji
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H.C
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Ajayi
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ABDUL
Practice Key Terms 3

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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